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Find all polynomials $P$ with integer coefficients such that $P(n)$ divides $2^n-1$ for all positive integers $n$.

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    $\begingroup$ Source of this problem, please? Partial results, if you know any? $\endgroup$ – Gerry Myerson Aug 15 '16 at 13:09
  • $\begingroup$ If there are infinitely many Mersenne primes, then all such polynomials $P$ must be constant. That is, $P\equiv -1$ or $P\equiv +1$. $\endgroup$ – Batominovski Aug 15 '16 at 13:15
  • $\begingroup$ Seems like either $P(n)\equiv +1\pmod{n}$ for all $n$ or $P(n)\equiv-1\pmod{n }$ for all $n$. $\endgroup$ – Batominovski Aug 15 '16 at 13:33
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    $\begingroup$ it's my own problem formulated while I was reading articles on Mersenne primes, I don't have any results (neither mine nor founded) $\endgroup$ – tong_nor Aug 15 '16 at 13:38
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    $\begingroup$ If one can show that the smallest prime divisors of Mersenne numbers are not polynomially bounded, then it is evident that $P$ is constant. However, I have not heard of any such bounds. $\endgroup$ – Batominovski Aug 15 '16 at 13:40
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Lemma: If $n | 2^n - 1$ for positive integer $n$, then $n = 1$. Proof: See here.

Main Proof:

Suppose $P(n) | 2^n - 1$ for all positive integers $n$. If $P(n)$ is constant, we know from the lemma that $P(n) = \pm 1$. Now suppose that $P$ is not constant. Thus,

$$P(x) = \sum_{k = 0}^j a_jx^j.$$

Now we claim that $a_0 \neq 0$. This follows directly from the lemma since $a_0 = 0$ implies $n|P(n) \implies n|2^n - 1$ for all integers $n$ which is a contradiction because only $n = \pm 1$ satisfy this condition from the lemma.

Now note that $$P(P(n) + n) = \sum_{k = 0}^j a_j(P(n) + n)^j \equiv \sum_{k = 0}^j a_jn^j \equiv 0 \pmod {P(n)}.$$

Thus, $P(n)$ divides $P(P(n) + n)$ and thus, divides $2^{P(n) + n} - 1.$ Therefore, $P(n)$ divides both of $2^n - 1$ and $2^{P(n) + n} - 1$ for all positive integers $n$.

Thus, $P(n)$ divides $2^{\gcd(n, P(n) + n)} - 1$. Now since $a_0$, the constant term of $P(n)$, is finite, it has finitely many prime factors. Pick $n$ to be one of infinitely many primes that don't divide $a_0$. Now if $n$ is a prime, $\gcd(n, P(n) + n) = n$ or $1$. It cannot be $n$ because $P(n) \equiv a_0 \pmod n$ and $n \nmid a_0$.

Thus, we can find infinitely many integers $n$ such that $\gcd(n, P(n) + n) = 1$. This means $$P(n)|2^{\gcd(n, P(n) +n)} - 1 \implies P(n) | 1$$ for infinitely many integers $n$.

Thus, $P(x)$ must be $\pm 1$ infinitely many times. By multiplying by a negative one if we need to, we must have $P(n) - 1$ has infinitely many roots which is a contradiction since the deg of $P$ is finite. Thus, only $P(x) = \pm 1$ work.

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  • $\begingroup$ You probably meant $n\nmid a_0$, rather than $p\nmid a_0$. Also, maybe it is worth mentioning how you use the lemma. I'm assuming that you used it to prove that $a_0\neq0$ (otherwise it wouldn't have finitely many prime divisors). Otherwise, I think the proof is perfect. $\endgroup$ – Batominovski Aug 15 '16 at 13:56
  • $\begingroup$ @Batominovski: Yup, Ill fix it. Thanks! (Also the post for the lemma is yours I believe.) $\endgroup$ – Sandeep Silwal Aug 15 '16 at 13:57
  • $\begingroup$ Yes, I proved the lemma. You got the same idea as I did (i.e., finding $m$ such that $P(n)$ divides $P(n+m)$) but I was unsuccessful at pursuing it. I thought about using the same lemma, but I didn't get as far as you. Bravo! $\endgroup$ – Batominovski Aug 15 '16 at 13:58
  • $\begingroup$ @Batominovski and Sandeep Do you know the source of this problem? It seems highly likely it was a contest problem. $\endgroup$ – Bill Dubuque Aug 15 '16 at 14:03
  • $\begingroup$ @Batominovski: Ok fixed all the errors I think. The main motivation for using $m = P(n)$ was that I wanted to use the gcd condition just like in the proof of the lemma and that choice made things a lot easier. $\endgroup$ – Sandeep Silwal Aug 15 '16 at 14:04

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